2 Find all the abelian groups of order 108 Do not list isomo

2. Find all the abelian groups of order 108. Do not list isomorphic groups.

Solution:

1. Here Z₁₀={0,1,2,...,9}, U(10) = {1,3,7,9}.

Since |Z₁₀| = 10 and U(10) = 4, the order of the group Z₁₀ (+) U(10) is = 10.4=40.

The element 3 belongs to Z₁₀ as order 10 and the element 3 belonges to U(10)

has order 4, so |(3,3)| = lcm(10,4) = 20.

So order of (3,3) in is 20.

2. Let G be a group of order 108.

Now 108 =2³ x3³.

The number of nonisomorphic abelian groups of order 108 is given by

p(2).p(3) where p(2) , p(3) are partitions of 2and 3 respectively.

Again p(2) = 2 as 2=1+1, 2=2 and p(3) = 3 as 3=1+1+1, 3=2+1, 3=3.

Therefore p(2).p(3) = 2x3=6.

These six groups are

Z₄XZ₂₇ , Z₄XZ₉XZ₃, Z₄XZ₃XZ₃XZ₃

Z₂XZ₂XZ₂₇, Z₂XZ₂XZ₉XZ₃, Z₂XZ₂XZ₃XZ₃XZ₃

3. Here \\phi : G\\to \\hbar {G} is a group homomorphism onto \\hbar {G} .

|G| =24 and |ker(\\phi)| = 3.

By first isomorphism theorem G/ker(\\phi) is isomorphic to \\hbar {G} .

Therefore |G| /|ker(\\phi)| = |\\hbar {G}|

or 24/3 = |\\hbar {G}|

or |\\hbar {G}| = 8.

Hence order of \\hbar {G} =8.